Thread: question ?

This second one doesn't requires too much thought to say it's unsolvable. Chess problems aren't the same.

That's absurd. I was really going to not post a reply, because it's obvious you're just trolling. However, this statement is just absurd.

First off, the problem is solvable within the stated requirements. Three solutions have been showen, all of which actually produce the required output in one form or another.

If the provided answers are "wrong", it's because the original poster's requirements were not spelled out clearly enough. Three answers meet the stated criteria.

"Chess problems aren't the same."

What? What exactly is a 'chess problem'? You make pretty big claims here and provide absolutely no backing what so ever.

Here's a "chess problem":
1) Show an ascii representation of a chess board in its initial state.
2) Black going first, make one move for each color.

There, you have a "chess problem", which is definately easier than the original question in this post. There is no need for anyone to actually post code, because this is a trivial "chess problem".

Again, your statement is absurd.

Quzah.

No more off topic posts please, they'll end up in the bin.

>>You didn't prototype func1 (or give it a return type)
It was prototyped (slightly incorrectly, but it was there). Look at the start of main():
>>int n, func1();
should really have been:
>>int n, func1(int);

>>m = t || exit(0); /*this is a short circuit operation using or*/
The use of the "or" (||) still constitutes a "test condition" which your requirements state is not allowed.

>>exit(0)
requires stdlib.h. Of course, if you include this header, the program won't compile at all, because, as Salem said, exit() returns void and therefore cannot be used as part of a conditional test.

>>recursion
As already stated, this constitutes a loop, and is also not allowed.

Code:

#include <stdio.h>
#include <math.h>

#define N 6
#define PI 3.14159265358979323846

int main()
{
	printf("%.2f\n",((N-1)*sin(0.00*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.05*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.10*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.15*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.20*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.25*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.30*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.35*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.40*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.45*PI))+1);
	printf("%.2f\n",((N-1)*sin(0.50*PI))+1);
	return 0;
}

This code prints out 11 numbers between 1 and N, inclusive. Define N in the code. (It wasn't specified that we had to print ALL the numbers between 1 and N)

what about that "recursive apporach" ???

A couple people have done it recursively. The clearest is AProg's, but he used an if, so it doesn't meet the criteria.

>>reall chees board
I had a reall chees board once, but I spelled it like this: "real cheese board" except for when I spelled it like this "etrfl keaess bard" It was this great big board, and it was made entirely out of cheese. There was some blue cheese, some colby cheese, some cheddar cheese, some swiss, motzeralla, and limburgher. All the cheese lived happily together until the cheddar proclaimed itself king and said "checkmate." AProg, try to make sense man, and stay on the subject. It's a cool problem and requires you to think. Don't dis it.